Answer
$\nu_2=606\ m/s$
$\dot{V_2}=2.74\ m³/s$
Work Step by Step
The energy balance is:
$\dot{m}(h_1+\nu_1^2/2)=\dot{m}(h_2+\nu_2^2/2)+\dot{Q}$
$h_1+\nu_1^2/2=h_2+\nu_2^2/2+\dot{q}$
From Table A-6:
Inlet (800 kPa, 400°C): $v_1=0.38429\ m³/kg,\ h_1=3267.7\ kJ/kg$
Outlet (200kPa, 300°C): $v_2=1.31623\ m³/kg,\ h_2=3072.1\ kJ/kg$
The mass flowrate is given by (with $A_1=0.08\ m²,\ \nu_1=10/m/s)$:
$\dot{m}=A_1\nu_1/v_1=2.082\ kg/s$
And given $\dot{Q}=25\ kW \rightarrow \dot{q}=12.008\ kJ/kg$
Solving the energy balance for the outlet velocity: $\nu_2=606\ m/s$
Since $\dot{V_2}=\dot{m}v_2$
$\dot{V_2}=2.74\ m³/s$
